J.R. S. answered 11/03/23
Ph.D. University Professor with 10+ years Tutoring Experience
Ca3(PO4)2(s) <==> 3Ca2+(aq) + 2PO43-(aq) ... Ksp = 1.99x10-29
Ksp = [Ca2+]3[PO43-]2
Let x = [PO43-] and then [Ca2+] = 1.5x
1.99x10-29 = (1.5x)3(x)2 = (3.375x3)(x2) = 3.375x5
x5 = 5.90x10-30
x = 1.43x10-6 M = [PO43-]
Molar solubility of Ca3(PO4)2 = 1/2 x 1.43x10-6 M = 7.15x10-7 M
For the solubility of Cu3(ASO4)2 in 0.325 M CuCl2, we need to consider the common ion effect. In this case, the common ion is Cu2+.
Cu3(ASO4)2(s) <==> 3Cu2+(aq) + 2AsO43-(aq)
Ksp = 7.57x10-36 = [Cu2+]3[AsO43-]2
7.57x10-36 = (0.325)3[AsO43-]2 = 0.0343 [AsO43-]2
[AsO43-]2 = 2.21x10-34
[AsO43-] = 1.49x10-17 M
Molar solubility of Cu3(ASO4)2 in 0.325 M CuCl2 = 1/2 x 1.49x10-17 M = 7.45x10-18 M
(be sure to check all the calculations)

J.R. S.
11/04/23